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What are the solutions to the system of equations?

What are the solutions to the system of equations?-example-1
User Vss
by
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2 Answers

3 votes

Answer:

(-1, 0) , (4.5, 11). (the second choice).

Explanation:

y = 2x^2 - 5x - 7

y = 2x + 2

Since both right side expressions are equal to y we can equate them.

2x^2 - 5x - 7 = 2x + 2

2x^2 - 7x - 9 = 0

(2x - 9)(x + 1)

x = 4.5 , -1.

Substituting these values of x in the second equation:

When x = -1 , y =2(-1) + 2 = 0.

When x = 4.5, y = 2(4.5) + 2 = 11.

User Melug
by
5.2k points
3 votes

Answer:

B

Explanation:

Given the 2 equations

y = 2x² - 5x - 7 → (1)

y = 2x + 2 → (2)

Since both equations express y in terms of x we can equate the right sides, that is

2x² - 5x - 7 = 2x + 2 ( subtract 2x + 2 from both sides )

2x² - 7x - 9 = 0 ← in standard form

Consider the factors of the product of the x² term and the constant term which sum to give the coefficient of the x- term

product = 2 × - 9 = - 18 and sum = - 7

The factors are + 2 and - 9

Use these factors to split the x- term

2x² + 2x - 9x - 9 = 0 ( factor the first/second and third/fourth terms )

2x(x + 1) - 9(x + 1) = 0 ← factor out (x + 1) from each term

(x + 1)(2x - 9) = 0

Equate each factor to zero and solve for x

x + 1 = 0 ⇒ x = - 1

2x - 9 = 0 ⇒ 2x = 9 ⇒ x = 4.5

Substitute these values into (2) for corresponding values of y

x = - 1 : y = (2 × - 1) + 2 = - 2 + 2 = 0 ⇒ (- 1, 0)

x = 4.5 : y = (2 × 4.5) + 2 = 9 + 2 = 11 ⇒ (4.5, 11)

Solutions are (4.5, 11) and (- 1, 0)

User Voscausa
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5.8k points